Mark Scheme (Results)

Summer 2017

Pearson Edexcel International A Level in Further Pure Mathematics F2 (WFM02/01)

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Summer 2017

Publications Code WFM02_01_1706_MS

All the material in this publication is copyright

© Pearson Education Ltd 2017

General Marking Guidance

All candidates must receive the same treatment. Examiners must mark

the first candidate in exactly the same way as they mark the last.

Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions.

Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie.

There is no ceiling on achievement. All marks on the mark scheme should be used appropriately.

All the marks on the mark scheme are designed to be awarded.

Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award

zero marks if the candidate’s response is not worthy of credit according to the mark scheme.

Where some judgement is required, mark schemes will provide the

principles by which marks will be awarded and exemplification may be limited.

When examiners are in doubt regarding the application of the mark scheme to a candidate’s response, the team leader must be consulted.

Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

EDEXCEL GCE MATHEMATICS General Instructions for Marking

1. The total number of marks for the paper is 75.

2. The Edexcel Mathematics mark schemes use the following types of marks:

M marks: method marks are awarded for ‘knowing a method and attempting to apply it’, unless otherwise indicated.

A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned.

B marks are unconditional accuracy marks (independent of M marks)

Marks should not be subdivided.

3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark

schemes.

bod – benefit of doubt ft – follow through

the symbol will be used for correct ft cao – correct answer only cso - correct solution only. There must be no errors in this part of the question to

obtain this mark isw – ignore subsequent working

awrt – answers which round to SC: special case oe – or equivalent (and appropriate)

dep – dependent indep – independent

dp decimal places sf significant figures The answer is printed on the paper

The second mark is dependent on gaining the first mark

4. All A marks are ‘correct answer only’ (cao.), unless shown, for example, as A1 ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd

answers should never be awarded A marks.

5. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected.

6. If a candidate makes more than one attempt at any question:

If all but one attempt is crossed out, mark the attempt which is NOT crossed

out. If either all attempts are crossed out or none are crossed out, mark all the

attempts and score the highest single attempt.

7. Ignore wrong working or incorrect statements following a correct answer.

General Principles for Further Pure Mathematics Marking

(But note that specific mark schemes may sometimes override these general principles).

Method mark for solving 3 term quadratic:

1. Factorisation

cpqqxpxcbxx where),)(()( 2, leading to x = …

amncpqqnxpmxcbxax andwhere),)(()( 2, leading to x = …

2. Formula

Attempt to use the correct formula (with values for a, b and c).

3. Completing the square

Solving 02 cbxx : 0,02

2

qcq

bx , leading to x = …

Method marks for differentiation and integration:

1. Differentiation

Power of at least one term decreased by 1. ( 1 nn xx )

2. Integration

Power of at least one term increased by 1. ( 1 nn xx )

Use of a formula

Where a method involves using a formula that has been learnt, the advice given in recent

examiners’ reports is that the formula should be quoted first.

Normal marking procedure is as follows:

Method mark for quoting a correct formula and attempting to use it, even if there are

small errors in the substitution of values.

Where the formula is not quoted, the method mark can be gained by implication from

correct working with values, but may be lost if there is any mistake in the working.

Exact answers

Examiners’ reports have emphasised that where, for example, an exact answer is asked

for, or working with surds is clearly required, marks will normally be lost if the candidate

resorts to using rounded decimals.

Answers without working

The rubric says that these may not gain full credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done “in your head”, detailed working would not be required.

Question

Number Scheme Notes Marks

1 2(cos0 isin0) or 2

(z =) 2 or (z =) 2(cos 0 + isin 0)

or 2cos0 + isin0 or 2 + 0i Allow 2(cos0 isin0 )

B1

2 22 cos isin

5 5

This answer in this form.

Do not allow e.g. 2

i52e

but allow 2 2

2cos 2isin5 5

B1

2 22 cos isin , ( 2,3,4)

5 5

k kk

Attempts at least 2 more solutions

whose arguments differ by2

5

. Allow

this mark if the arguments are out of

range. May be implied by their

answers.

M1

Note that this answer in general solution form can score full marks if

correct i.e. the A marks below can be implied.

E.g. 2 2

2 cos isin , ( 0, 1, 2, 3, 4)5 5

k kz k

scores full marks

4 42 cos isin

5 5

6 62 cos isin

5 5

8 82 cos isin

5 5

A1: One further correct answer, allow

the brackets to be expanded.

A1 A1

A1: All correct, allow the brackets to

be expanded.

Do not allow 4 4

2 cos isin5 5

or

4 42 cos isin

5 5

for 6 6

2 cos isin5 5

Do not allow 2 2

2 cos isin5 5

or

2 22 cos isin

5 5

for 8 8

2 cos isin5 5

Ignore answers outside the range.

For a fully correct solution that has extra solutions in range, deduct the final A mark.

Answers in degrees: Penalise once the first time it occurs.

Answers in degrees are: 0, 72, 144, 216, 288

(5)

Total 5

Question

Number Scheme Notes Marks

2. 4 5

( 3) ( 3)

x

x x x

Way 1

4 5

0( 3) ( 3)

x

x x x

Collects expressions to one side M1

2 4 5

0( 3)

x x

x x

M1: Attempt common denominator M1A1

A1: Correct single fraction

0, 3x Correct critical values B1

2 4 5 ( 5)( 1) 0

...

x x x x

x

Attempt to solve their quadratic as far as

x = … to obtain the other 2 critical

values

M1

1,5x Correct critical values A1

3 1, 0 5

or e.g.

( 3, 1] (0, 5]

x x

M1: Attempts two inequalities using

their 4 critical values in ascending order.

E.g. * * , * *a x b c x d where * is

“< “ or “≤” and a < b < c < d or

equivalent inequalities. Dependent on at

least one earlier M mark.

dM1A1A1

A1: All 4 cv’s in the inequalities correct

A1: Both intervals fully correct

Notes

Intervals may be separated by commas, written separately, or “or” or “and” may be used but not

All marks are available for correct work if “=” is used instead of " " for the first 6 marks

(9)

Total 9

Way 2

22Multipliesby 3x x

2 3 4 5 3x x x x x

22Multipliesbothsidesby 3x x .

May multiply by more terms but must be

a positive multiplier containing

22 3x x

M1

3 2( 3) 4 ( 3) 5 ( 3) 0x x x x x x M1: Collects expressions to one side M1A1

A1: Correct inequality (or equation)

0, 3x Correct critical values B1

( 3)( 5)( 1) 0 ...x x x x x

Attempt to solve their quartic as far as

x = … to obtain the other 2 critical

values. Allow the x and x + 3 to be

“cancelled” to obtain the other critical

values so may end up solving a cubic or

even a quadratic.

M1

1,5x Correct critical values A1

3 1, 0 5

or e.g.

( 3, 1] (0, 5]

x x

M1: Attempts two inequalities using

their 4 critical values in ascending order.

E.g. * * , * *a x b c x d where * is

“< “ or “≤” and a < b < c < d or

equivalent inequalities. Dependent on at

least one earlier M mark.

dM1A1A1

A1: All 4 cv’s in the inequalities correct

A1: Both intervals fully correct

(9)

Way 3

Draws a sketch of graphs

4

3

xy

x

and

5

3y

x x

0, 3x Correct critical values (vertical

asymptotes) B1

4 5

( 3) ( 3)

x

x x x

Eliminate y M1

4 5x x M1: Obtains quadratic equation M1A1

A1: Correct quadratic equation

2 4 5 0 1,5x x x M1: Solves their quadratic equation as

far as x = … M1A1

A1: Correct critical values

3 1, 0 5

or e.g.

( 3, 1] (0, 5]

x x

M1: Attempts two inequalities using

their 4 critical values in ascending order.

E.g. * * , * *a x b c x d where * is

“< “ or “≤” and a < b < c < d or

equivalent inequalities. Dependent on at

least one earlier M mark.

M1A1A1

A1: All 4 cv’s in the inequalities correct

A1: Both intervals fully correct

If the candidate takes the above approach and there is no sketch e.g. just cross multiplies to obtain the critical

values -1 and 5 then no marks are available i.e. the cv’s 0 and -3 must be stated somewhere to give access to

subsequent marks in this case.

Considers Regions:

Way 4

Considers 3 3 0x x x

4 5 1 5x x x

But 3x so no solution

Can be marked as:

B1: Critical values 0 and -3

M1: Considers 3 regions

M1: Obtains quadratic equation

A1: Correct quadratic

M1: Solves quadratic

A1: cv’s -1 and 5

Final 3 marks as already defined.

Considers 3 0 3 0x x x

4 5 5 or 1x x x x

But 3 0x so 3 1x

Considers 0 3 0x x x

4 5 1 5x x x

But 0x so 0 5x

Question

Number Scheme Notes Marks

3.(a)

3 3 3 3 2

2 33 3 2

( 1) ( 3 3 1)

3 3or 1 1 1

1 2

r r r r r r

r r r r

23 3 1r r *

or 3 3 2 2( 1) ( ( 1) ( 1) )r r r r r r

23 3 1r r *

Shows a correct expansion of 3( 1)r or

uses 3 3 2 2( )( )a b a b a ab b

and achieves the printed answer with

no errors.

B1*

(1)

(b) 3 3( 1)n n 3 3( 1) ( 2)n n 3 3( 2) ( 3)n n

…

… 3 33 2

3 32 1 3 31 0

Uses the method of differences. Must

include at least 1,2,....,r n or

1,...., 1 , .r n n But may implied

by sight of 3 3 3( 1)r r n if

insufficient terms shown. If method is

clearly other than differences (see note

below), then score M0.The final A

mark can be witheld if differences not

shown i.e. just writes down n3.

M1

3 2 2

1 1 1 1

3 3 1 3 3 1n n n n

r r r r

n r r r r

Sets 3 2

1

3 3 1n

r

n r r

and attempts to expand RHS

M1

1

1n

r

n

1

1n

r

n

seen or implied B1

2 32

1

3 ( 1)( 1) ( 1)n

r

r n n n n n

Rearranges to make 2

1

n

r

k r

the

subject and substitutes for 1

n

r

r

.

Dependent on the first method

mark.

dM1

2

1

1( 1)(2 1)

6

n

r

r n n n

** Completely correct solution with no

errors seen. A1*

Allow e.g.

3 22 3 1( 1)(2 1)

6 6

n n nn n n

(5)

Note: May be seen in (b):

3 2 23 2 2 3

1

1 11 1 1

4 4

n

r

r r n n n n n

etc.

Scores a maximum M0M1B1dM0A0 (not using differences)

Generally, there are no marks for proof by induction

Total 6

Question

Number Scheme Notes Marks

4(a)

3e cos3 e sin 3

d3e cos3 9e sin 3 e sin 3 3 e cos3

d

( 3 3 )e cos3 ( 9 )e sin 3

x x

x x x x

x x

y x A x

yx x A x A x

x

A x A x

Attempts to differentiate the given expression by using the product rule on

3e cos3x x to give e cos3 e sin3x xx x or by using the product rule on

e sin3xA x to give e cos3 e sin3x xA x A x

M1

2

2

d( 24 6 )e cos3 (18 8 )e sin3

d

x xyA x A x

x

(Terms may be uncollected)

Uses the product rule again on an expression of the form e sin3x x or e cos3x x to

give e cos3 e sin3x xx x . Dependent on the first method mark.

dM1

2

2

d d2 10 (12 12 )e cos3 (36 4 )e sin3

d d

x xy yy A x A x

x x

Substitute their results into the differential equation. (May be implied)

M1

12 12 0A or 36 4 40 ...A A

Compares coefficients of

e sin3x x or e cos3x x and

attempts to find A. Dependent on

the previous method mark.

dM1

1A cao A1

(5)

(b)

Marks

for (b)

can score

anywhere

in their

answer.

2 2 10 0 1 3im m m

M1: Forms and attempts to solve

the Auxiliary Equation. See

General Principles. M1 A1

A1: Correct solution for the AE

e ( cos3 sin3 )xy C x D x

or

(1 3i) (1 3i)e ex xy C D

Correct form for CF using their

complex roots from the AE M1

e ( cos3 sin3 ) 3e cos3 e sin3x x xy C x D x x x

GS = their CF + their PI (Allow ft on their CF and PI )

Must start y = … and depends on at least one the M’s being scored and must have

been using a PI of the form given.

A1ft

(4)

(c) 0, 3 3 3 0x y C C Attempts to substitute x = 0 and

y = 3 into their answer to (b) M1

d( 3 )e cos3 ( 3 )e sin3 10e sin3

d

x x xyC D x C D x x

x

Attempt to differentiate their GS with or without their C

M1

3 3C D

Attempt to substitute x = 0 and

d3

d

y

x into their

d

d

y

x M1

e sin3 3e cos3 e sin3x x xy x x x Correct answer. Must start y = … A1cao

(4)

Total 13

Question

Number Scheme Notes Marks

5 2cos=e xy

(a) 2 2cos cosd= 2sin cos e sin 2 e

d

x xyx x x

x

M1: Differentiates using the chain rule to obtain an expression of the form 2cossin cos e xx x or

2cossin 2 e xx

A1: Correct derivative

21Note that candidates mayuse 1 cos 2 insteadof cos

2x x

M1A1

2 22

cos cos

2

d= sin 2 ( 2sin cos e ) 2cos 2 e

d

x xyx x x x

x

Correct use of the Product Rule on their first derivative

Dependent on the first method mark.

dM1

22

cos 2

2

de (sin 2 2cos 2 )

d

xyx x

x * Achieves the printed answer with no

errors. A1*

Alternative using logs:

2cos 2 1 d=e ln cos 2sin cos

d

x yy y x x x

y x

M1: 1 d

sin cos or sin 2d

yk x x k x

y x A1:

1 d2sin cos

d

yx x

y x

2

2

d d dsin 2 sin 2 2 cos 2

d d d

y y yy x x y x

x x x

M1: Correct use of product rule

22

cos 2

2

de (sin 2 2cos 2 )

d

xyx x

x *

A1: Achieves the printed answer with no errors.

(4)

(b) 2

2

d d0, 2e

d d

y y

x x Both seen, can be implied by

subsequent work. B1

2 2 2

2

cos 0 cos 0 cos 0 2 2

f f 0 f 0 f 0 ...2

1e sin 0e e (sin 0 2cos0) ...

2

xx x

x x

Applies the correct Maclaurin expansion, the 1

" "2

is required and there must

be no x’s in the derivatives.

This can be implied by their expansion but if the expansion in incorrect for

their values and the formula is not quoted, score M0.

M1

2cos 2e e(1 ...)x x

Or exact equivalent e.g. 2e e x

(i.e. all trig. evaluated) A1

(3)

Total 7

Question

Number Scheme Notes Marks

6. 2dcos sin (cos ) ln

d

yx y x x x

x

d sincos ln

d cos

y xy x x

x x

Attempt to divide through by

cos x. If the intention is not clear,

must see at least 2 terms divided by

cos x.

M1

sind

lncoscose ex

xxxI

M1: their P( ) d

ex x . Dependent on the

first method mark. dM1A1

A1: lncos lnsece orex x

1

cos x 11

or (cos ) or seccos

x xx

A1

ln dcos

yx x

x

or

dln

d cos

yx

x x

M1: their their dy I Q x I x or

d

their theird

y I Q x Ix

M1A1 A1: ln d or

cos

yx x

x

dln

d cos

yx

x x

lncos

yx x x C

x

Attempts ln dx x by parts correctly

(correct sign needed unless correct

formula quoted and used).

M1

( ln )cosy x x x C x Any equivalent with the constant

correctly placed and “y = …” must

appear at some stage.

A1

Total 8

Note: Failure to divide by cos x at the start would mean that only the 3rd

Method mark is available.

Question

Number Scheme Notes Marks

7

(a) sin 4cos2 siny r

Attempts to use sinr M1

d4cos 2 cos 8sin 2 sin

d

y

or

2 3 2d4 1 2sin sin 4sin 8sin 4cos 24sin cos

d

yy

A correct expression for d

d

y

or any multiple of

d

d

y

B1

d0 ...

d

y

Set their d

0d

y

and attempt to solve to

obtain a value for θ

M1

8, 0.421, 2.72

3r Any one of:

8(or awrt 2.7)

3

or 0.421... or 2.72...

r

A1

8

3

0.421, 2.72

r

Correct value for r and both angles correct.

May be seen as 8 8,0.421 , ,2.72

3 3

. Allow

8 80.421, , 2.72,

3 3

but coordinates do not

have to be paired and accept awrt 0.421, 2.72

and allow awrt 2.7 for 8

.3

Ignore any other

coordinates given once the correct values

have been seen.

A1

(5)

(b) 2... (4cos2 ) dA

Indication that the integration of 2(4cos2 )

is required. Ignore any limits and ignore any

constant factors at this stage.

M1

2 1cos 2 (1 cos 4 )

2 A correct identity seen or implied. A1

... sin 4A

Integrates to obtain an expression of the form

sin 4 . Ignore any limits and ignore

any constant factors. Dependent on the first

method mark.

dM1

4

0

116 sin 4

4

A fully correct method that if evaluated

correctly would give the answer 4 . Note

that the correct “constant factor” may only be

applied at the very last stage of their working

and this method mark would only be awarded

at that point. Dependent on all previous

method marks.

ddM1

Examples that could score the final M1 (following correct work):

4 4 2

30 0

4 4

1 1 1 116 sin 4 , 8 sin 4 , 8 sin 4 , 16 sin 4

4 4 4 4

4 cao A1

(5)

(c)

162 sin

3 6PQ r

Correct expression or value for PQ or PQ/2.

8 1 8E.g. 2 , 2 sin 0.421,

3 36

8 8 62 sin 2.72, or half of these.

3 9

May be implied by awrt 2.2 or awrt 1.1

B1

8SP or 42

SP

Correct value for SP or SP/2 B1

16 64 6Area 8

93 6PQRS

Their .PQ SP Must be the complete

rectangle here. M1

128Required area 4

3 6

M1: Their rectangle area – their answer to

part (b)

M1A1 A1: Correct exact answer or equivalent exact

form e.g. 64 6

49

or allow awrt 4.8 or 4.9

(5)

Total 15

Question

Number Scheme Notes Marks

8(a)(i) 5 5 4 3 2 2 2 3 3 4 4 5 5cos5 isin5 ( i ) 5 i 10 i 10 i 5 i ic s c c s c s c s c s s

Attempts to expand 5( i )c s including binomial coefficients (NB may only see real

terms here)

M1

5 5 3 2 2 4 4 5 3 2 4cos5 Re( i ) 10 i 5 i 10 5c s c c s c s c c s cs

Extracts real terms and uses i2 = -1 to eliminate i. M1

5 3 2 4cos5 cos 10cos sin 5cos sin *

Achieves the printed result with no errors seen. A1*

Alternative:

1cos isin , cos isin , cos isinnz z z n n

5

5 4 3 2

2 3 4 5

1 1 1 1 1 15 10 10 5z z z z z z

z z z z z z

5

2cos 2cos5 10cos3 20cos

M1: Expands

51

zz

including binomial coefficients and uses

12cosn

nz n

z

at least once to obtain an equation in cos 5cos5 16cos 5cos3 10cos

3 5 3cos3 4cos 3cos cos5 16cos 20cos 15cos 10cos

M1: Uses correct identity for cos3θ to obtain cos5 in terms of single angles

2

5 2 3cos 15cos 1 sin 20cos 5cos

5 2 4cos 10cos sin 15cos sin

5 4 2 2cos 5cos sin 10cos sin sin 1

5 3 2 4cos5 cos 10cos sin 5cos sin *

A1: Achieves the printed result with no errors seen (may need careful checking)

(ii) 4 2 3 5sin5 5cos sin 10cos sin sin

This expression (or equivalent) with no i’s seen.

Note that some candidates may re-start and expand here as above.

B1

(4)

(b) 4 2 3 5

5 3 2 4

sin5 5cos sin 10cos sin sintan5

cos5 cos 10cos sin 5cos sin

Uses sin 5

tan 5cos5

and substitutes the results from part (a)

M1

3 5 5 3

2 4 4 2

5tan 10 tan tan 10 5*

1 10 tan 5tan 5 10 1

t t t

t t

Achieves the printed result with no errors seen.

A1*

Note that a minimum could be: 4 2 3 5 5 3

5 3 2 4 4 2

5cos sin 10cos sin sin 10 5tan5 *

cos 10cos sin 5cos sin 5 10 1

t t t

t t

Note that some candidates may work backwards which is acceptable:

E.g.

5 3 5 3

4 2 4 2

10 5 tan 10 tan 5tan

5 10 1 5tan 10 tan 1

t t t

t t

4 2 3 5

5 3 2 4

5cos sin 10cos sin sintan5

cos 10cos sin 5cos sin

(2)

Pearson Education Limited. Registered company number 872828 with its registered office at Edinburgh Gate, Harlow, Essex CM20 2JE

(c)

5 3

4 2

5 3

4 2

tan 10 tan 5 tantan 5 0 or 0

5 tan 10 tan 1

10 5or 0

5 10 1

t t t

t t

Considers tan5 0 .

This may be implied by 5 3

5 3

tan 10 tan 5tan 0

or 10 5 0 ort t t

4 2

4 2

tan 10 tan 5 0

or 10 5 0t t

M1

5 3

5 3

tan 10 tan 5tan 0

or 10 5 0t t t

Equate numerator to 0

This may be implied by 4 2

4 2

tan 10 tan 5 0

or 10 5 0t t

M1

4 2

4 2

tan 10 tan 5 0

or 10 5 0t t

Correct quartic A1

2 10 5 0x x 2 10 5 0x x or equivalent A1

(4)

(d) Product of roots: 2 2 2

tan tan 55 5

Or solves 2" 10 5 0"x x and attempts to

multiply roots together e.g.

10 100 205 2 5

2x

and

5 2 5 5 2 5 ...

Must clearly state product of roots

or e.g. αβ = 5 or x1x2 = 5 and uses

their constant in (c) or solves their

quadratic and attempts product of

roots.

M1

2 2 2 2tan tan 5 tan tan 5

5 5 5 5

* Shows the given result with no

errors. A1

(2)

Total 12